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(C) Let the speed of the bus be $v_B$ and the speed of the cyclist be $v_C = 20 \,km/h$.Let $d$ be the distance between two consecutive buses,where $d

Vedclass · Accepted Answer

$\frac{3}{20} \,h$ and $40 \,km/h$

Vedclass · Answer

(C) Let the speed of the bus be $v_B$ and the speed of the cyclist be $v_C = 20 \,km/h$.Let $d$ be the distance between two consecutive buses,where $d = v_B 	imes T$.Case $1$: Bus moving from $A$ to $B$ (same direction as cyclist).The relative speed of the bus with respect to the cyclist is $(v_B - v_C)$.The time interval between buses passing the cyclist is $t_1 = 18 \,min = \frac{18}{60} \,h = 0.3 \,h$.Thus,$d = (v_B - 20) 	imes 0.3 \dots (i)$.Case $2$: Bus moving from $B$ to $A$ (opposite direction to cyclist).The relative speed of the bus with respect to the cyclist is $(v_B + v_C)$.The time interval between buses passing the cyclist is $t_2 = 6 \,min = \frac{6}{60} \,h = 0.1 \,h$.Thus,$d = (v_B + 20) 	imes 0.1 \dots (ii)$.Equating $(i)$ and $(ii)$:$(v_B - 20) 	imes 0.3 = (v_B + 20) 	imes 0.1$$3(v_B - 20) = v_B + 20$$3v_B - 60 = v_B + 20$$2v_B = 80 \Rightarrow v_B = 40 \,km/h$.Now,substitute $v_B$ into $d = v_B 	imes T$ using equation $(ii)$:$v_B 	imes T = (v_B + 20) 	imes 0.1$$40 	imes T = (40 + 20) 	imes 0.1$$40T = 60 	imes 0.1 = 6$$T = \frac{6}{40} \,h = \frac{3}{20} \,h$.

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